CEE 3310 – Open Channel Flow,, Nov. 18, 2016 165

8.1 Review

Drag & Lift ⇒ Laminar vs Turbulent Boundary Layer ⇒ Turbulent boundary

layers stay attached to bodies longer ⇒ Narrower wake! ⇒ Lower pressure drag!

CD =FD

12ρV 2A

whereA is the appropriate area (i.e., surface, cross− sectional, wetted)

CD = f(Re) – may have to iterate to get a converged solution

8.2 Open-Channel Flow

Pipe/duct flow ⇒ closed, full, gas or liquid.

⇒ Pressure – friction balance.

Open-channel flow ⇒ free-surface (river, flume, partially filled pipe, ...).

⇒ Gravity – friction balance.

Open-channel flow is often turbulent with complex geometries (rivers, estuaries, streams!).

Hence we will simplify the geometry (often an excellent approximation)!

⇒ We will assume straight channels with simple geometries (prismatic channels) and

steady state flow (in time).

Free-Surface ⇒ at constant pressure ⇒ atmospheric.

This helps analysis! But free-surface position is unknown! This hinders analysis!

There are entire books written on open-channel flow (French, Henderson, Chow, ...), we

will only discuss the basics. For more details keep an eye open for CEE 3320 (may get a

4000 number) that will cover the details of unsteady open channel/river flow, sediment

transport, and fundamental coastal engineering/water wave theory.

c© 2016 Edwin A. Cowen

CEE 3310 – Open Channel Flow,, Nov. 18, 2016 166

8.3 1-D Flow Assumption

Consider the energy equation. At surface P1 = P2 = Patm, therefore:

V 21

2g+ z1 =

V 22

2g+ z2 + hf

From our pipe flow work we should immediately suspect that it is a reasonably good

approximation to consider:

hf ≈ fL

Dh

V 2avg

2g

where L = x2−x1 and Dh = 4A/P as previously defined. Note P is the wetted perimeter

hence only the distance along the wetted sides of the channel.

Now, we can relate V1 to V2 from conservation of mass since Q1 = Q2 ⇒ V2 = V1A1

A2

.

We define an appropriate local Reynolds number as ReDh= UDh/ν. If ReDh

>∼ 105 the

flow is turbulent and most are (classic exception is sheet flow of paved surfaces).

c© 2016 Edwin A. Cowen

CEE 3310 – Open Channel Flow,, Nov. 18, 2016 167

8.4 Flow Classification by Variation of Depth with

Distance Along Channel

• Uniform Flow constant depth ⇒ dydx

= 0.

• Varied Flow non-constant depth ⇒ dydx6= 0.

– Gradually Varied Flow ⇒ dydx� 1.

– Rapidly Varied Flow ⇒ dydx∼ 1.

A picture:

8.5 Uniform Flow

In uniform flow y1 = y2, V1 = V2 = V , therefore the energy equation becomes:

z1 − z2 = S0L = hf

Flow is essentially fully developed, therefore we can apply Darcy-Weissbach relations.

hf = fL

Dh

V 2avg

2g

In open channel flow it is more common to work with the:

hydraulic radius =A

P=Dh

4= Rh

c© 2016 Edwin A. Cowen

CEE 3310 – Open Channel Flow,, Nov. 18, 2016 168

where again (P) is the length of the wetted perimeter. Combining the above two equa-

tions we have:

S0L = fL

Dh

V 2avg

2g⇒ V =

(8g

f

)1/2

(RhS0)1/2

8.6 Chezy Formulas

Chezy defined the coefficient

C =

(8g

f

)1/2

(now called the Chezy coefficient) and found that it varies by a factor of 3. Therefore

V = C (RhS0)1/2

and

Q = CA (RhS0)1/2

Manning did field tests and found

C =

(8g

f

)1/2

≈ αR

1/6h

n

where n is known as Manning’s n and is a roughness coefficient and α is a dimensional

constant that varies with systems of units (this is not a homogeneous equation, remem-

ber?!). For SI α = 1, for BG α = 1.486. It is left as an exercise for the student to find

the units and verify the conversion.

8.7 Manning’s Equation

Substituting Manning’s result into the Chezy formulas we have the celebrated Manning’s

equation:

V ≈ α

nR

2/3h S

1/20 and Q ≈ α

nAR

2/3h S

1/20

n varies by a factor of 15 and is tabulated in your text and more extensively elsewhere.

c© 2016 Edwin A. Cowen

CEE 3310 – Open Channel Flow,, Nov. 18, 2016 169

8.8 Example – Fall Creek Flow

Let’s consider Fall Creek where the USGS operates a gaging station. Looking over

yesterday’s record we see a local maximum flow depth occurred on 11/17/2016 @ 20:45

EST of d = 0.85′. If we estimate S0 ≈ 0.001, n = 0.035 from table 10.1 in text,

assuming somewhere between clean and straight and sluggish with deep pools and a width

of b = 50′, d = 0.85′ we can ask the question, what is Q?

Q =α

nAR

2/3h S

1/20

Rh =A

P=

bd

2d+ b≈ d if b� d

therefore Q =α

nbd5/3S

1/20 =

1.486 ft1/3/s

0.035(50 ft)(0.85 ft)5/3

√0.001

⇒ Q = 51 CFS

Actual value from calibrated flow gage was 53 CFS. Just using ball park estimates we

were within 4% – amazingly good (best I have ever seen)! Hard to be more accurate as

with luck we pretty much hit it but in reality one would need a survey of the river slope

and the wetted perimeter. We could get these more accurately from a USGS topographic

map but to be truly accurate we would send a survey team out to measure directly in

the field.

c© 2016 Edwin A. Cowen